- Let $X:\Omega\to \mathbf{X}$ be an integrable random variable where $\mathbf{X}\subset \mathbb{R}^d$ is a compact set.
- Denote $\mathbf{X}_n = \{\mathbf{X}_n(1),\dots, \mathbf{X}_n(n)\}$ to be a n-component partition of $\mathbf{X}$ such that partition $\mathbf{X}_{n+1}$ is a refinement of $\mathbf{X}_n$.
- Set $x_{n,i} \in \mathbf{X}_n(i)$ for all $n\in\mathbb{N}$ and $i\leq n$.
- Define random variable $X_n = \sum_{i=1}^n x_{n,i} \mathbb{1}(X \in \mathbf{X}_n(i))$ where $\mathbb{1}(\mathbf{B})$ is the indicator function for the set $\mathbf{B}$.
What conditions do I need for $X_n$ to converge in distribution to $X$???
How about if the sigma-alegbras generated by $ \sigma^{(n)} =\sigma(\{X \in \mathbf{X}_{(n)}(i)\}, \enspace i=1, \dots, n)$ satisfies $\sigma(X) = \sigma(\cup_{n\in\mathbb{N}}\sigma^{(n)})$? Is this sufficient?
A reasonable condition is that $$ \tag{*} \lim_{n\to +\infty}\max_{1\leqslant i\leqslant n}\operatorname{diam}\left(\mathbf X_n(i)\right)=0, $$ where $\operatorname{diam}\left(A\right)=\sup_{x,y\in A}\left\lVert x-y\right\rVert$. In this case, since $$ X_n\left(\omega\right)-X\left(\omega\right) =\sum_{i=1}^n\left(x_{n,i}-X\left(\omega\right)\right) \mathbf 1\left\{X\left(\omega\right)\in \mathbf X_n(i)\right\}, $$ we have $$ \left\lvert X_n\left(\omega\right)-X\left(\omega\right)\right\rvert \leqslant \max_{1\leqslant i\leqslant n}\operatorname{diam}\left(\mathbf X_n(i)\right). $$ If $(*)$ is not satisfied, then there is no reason to have the convergence of $X_n$ to $X$: for example, choose $d=1$, $\mathbf X=(0,2)$ and consider $\mathbf X_{n,1}=(0,1]$ for each $n$ and any sequence of finer and finer partitions of $(1,2)$.